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STRUCTURE OF Sb-121 AND Sb-123
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) ( July 2014) Unfortunately the discovery of the assumed uncharged neutron (1932) along with the invalid relativity led to the abandonment of electromagnetic laws in favor of wrong theories which could not lead to the nuclear structure. Under this physics crisis in 2003 I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” by reviving the natural laws which led to my discovery of 288 quarks in nucleons including 9 charged quarks in proton and 12 ones in neutron able to give the nuclear binding and nuclear structure by applying the well-established laws of electromagnetism. (See my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS ). Nuclear structure of Sb-103 of S = +5/2 with 26 blank positions Antimony (Sb) occurs in two stable isotopes, 121Sb and 123Sb. Since the structure of Sb-102 cannot exist we compare the Sb-102 of 51 protons and 51 neutrons(odd number ) with the Sn-100 of 50 protons and 50 neutrons (even number) . Under this comparison we present the structure of the well-known Sb-103 with S = +5/2 which breaks the high symmetry of Sn-100 ( See my STRUURE OF Sn-112...Sn-124 ). Since the Sn-100 of high symmetry is elongated with the two symmetrical squares, here the additional n51p51 with the p50n50 of Sn-100 with S = 0 make the fourth alpha particle with S = 0 existing behind the central parallelepiped. Moreover the n39p39 of the square of the Sn-100 changes the spin from S =-1 to S =0 giving S = +1 in order to make the alpha particle with S =0. Here it is moved to make with p49n49 the third symmetrical alpha particle existing in front of the central parallelepiped. Under this condition the n37p37 of the square of Sn-100 with S = 0 is moved to fill the blank positions existing in front of p38n38 .So it changes the spin from S =-1 to S =+1 giving S = +2. Thus under the S = +3 the additional n52(-1/2) gives he structure of Sb-103 with S = +5/2. That is S = +3 + 1(-1/2) = +5/2 In other words these three deuterons of n37p37, p38n38 and n40p40 of the seventh horizontal plane in Sb-103 with S = +5/2 reduce the 8 horizontal planes to 7 ones in order to decrease the pp repulsions of long range. Of course this situation here contributes to the reduction of 8n of the squares to 2n. However the change of two rectangles into the two alpha particles decreases the 4(n) to 2(n) and increases the 2n to 4n. Under this condition the number N of blank positions is given by The change of squares with 2n The first and sixth plane with 4(n) The second and fifth plane with 4{n} +8n The third and fourth plane with 8(n) That is N = 2n + 4(n) + 4{n} + 8n + 8(n) 26 extra neutrons of opposite spins Or N = 4{n} + 10n + 12(n) = 26 extra neutrons of opposite spins ' ' STRUCTURE OF Sb-121 WITH S = +5/2 ''' After a careful analysis I found that the structures of the above stable nuclide is based on the structure of Sb-103 with S = +5/2 beause the 18 extra neutrons than the one n52 of Sb-102 have opposite spins giving S =0. '''STRUCTURE OF Sb-123 WITH S = +7/2 In the same way the stable structure of Sb-123 of S = +7/2 with 20 extra neutrons than the one n52 of Sb-103 is based on the same structure of of Sb-103 with S =+5/2. Here it has two more extra neutrons than the one n52 of Sb-103 and 18 extra neutrons of opposite spins giving S = 0. That is S = +5/2 + 2(+1/2) + 0 = +7/2 DIAGRAM OF Sb -103 WITH S = +5/2 FORMING 22 BLANK POSITIONS Here you see the p47n47 along with the p48n48, which make two symmetrical alpha particles of opposite spins . But you cannot see the additional p49n49 the n39p39 of the third alpha particle and the n50p50 and the p51n51 of the fourth alpha particle. Also the p41, n41, p42, n42, p43, n43, p44, and n44 which form the central parallelepiped of opposite spins are not shown. Also the 8 deuterons of opposite spins from p13n13 to p20n20 and the 4 deuterons from p33n33 to p36 n36 are not shown. Since the change of spins of p37n37 and n39p39 gives S = +3 we get the structure of Sb-103 by adding the np2(-1/2), which is not shown. Moreover the extra neutrons 4(n) of the first and the sixth plane are not shown, while the extra neutrons 2n existing over the p31 and p32, with the extra 4n existing near the p23, p24, p29 and p30, are shown. ' n40.......p40.......n ' ' n........p38..........n38 Seventh H. plane ' ' n31………p12.........n12.......p32' ' p31........n11.........p11…… n32 Sixth H. plane' ' n........ p29.........n10.........p10…… n30' ' n29………..p9..........n9 …….p30..........n Fifth H. plane' ' p47.......n27.........p8..........n8.........p28......... n48' ' n45.......p27.........n7..........p7........n28..........p46 Fourth H. plane' ' n47......p25.........n6.........p6..........n26...........p48' ' p45......n25……….p5..........n5……….p26.........n46 Third H. plane' ' n23………p4........n4………….p24...........n' ' n........p23……..n3………p3………..n24 Second H.plane' ' p21.........n2………p2............n22' ' n21........p1........n1.........p22 First H. plane' ' ' ' ' TOP VIEW OF THE FIRST HORIZONTAL PLANE IN WHICH ALL NUCLEONS ARE SHOWN ' HERE THE FIRST EXTRA NEUTRON (n ) MAKES THE TWO RADIAL BONDS WITH p22 AND p33 WHILE THE SECOND ONE MAKES THE TWO RADIAL BONDS WITH p21 AND p34 ' (n)........p34....... n34 ' p21....... n2........ p2....... n22 ' ' n21.........p1. .......n1.......p22 ' ' n33.......p33..... (n)' TOP VIEW OF THE SECOND HORIZONTAL PLANE Here the n near the p14 fills the blank position formed by p51 and p14. While the {n} near the p14 fills the blank position formed by p14, p24 and p44. The 2n near p24 and p23 fill the blank positions formed by p24 and p48 as well as by p23 and p45. Moreover the blank position of {n} near p23 is formed by p23, p13 and p41. Finally the blank position of n near the p13 is formed byp13 and p49. That is we have 2{n} +4n and the same situation occurs at the fifth horizontal plane. ' n' ' n14.......p14........{n}' ' n23.......p4........n4.........p24......n' ' n.......p23........n3........p3.........n24' ' {n}.......p13......n13' ' n ' ' ' ' ' TOP VIEW OF THE THIRD HORIZONTAL PLANE WITH POSITIVE SPINS ' '''HERE YOU SEE THE p41, n43, n42 AND p44 WHICH MAKE THE SQUARE OF THE CENTRAL PARALLELEPIPED. YOU SEE ALSO THE p45 AND n47 ALONG WITH THE n46 AND p48 WHICH MAKE THE SYMMETRICAL SQUARES OF THE TWO ALPHA PARTICLES. AT THE SAME PLANE ARE SHOWN ALSO THE DEUTERONS LIKE n16p16 AND p15n15 ALONG WITH THE ADDITIONAL p49, n39, n50, AND p51 WITH 4(n) ' n50.......p51......(n) ' ' n42........p16......n16......p44.........(n)' ' n47........p25........n6........p6........n26.........p48''' ' p45........n25........p5........n5........p26........ n46' ' (n)..........p41.......n15.......p15.......n43' ' (n)........p49.......n39' ' ' Category:Fundamental physics concepts